Let $\mathbb{D}$ denote the open unit disk in $\mathbb{C}$, let $\mathcal{A}\left(\mathbb{D}\right)$ denote the vector space of all complex-valued functions which are holomorphic on $\mathbb{D}$, and let $W$ (“the set of weights”) be the set of all continuous non-negative real-valued functions on $\left[0,1\right]\subseteq\mathbb{R}$ that vanish at $1$.

Now, for every $w\in W$, define: $${A}_{w}\left(\mathbb{D}\right)\overset{\textrm{def}}{=}\left\{ f\in\mathcal{A}\left(\mathbb{D}\right):\limsup_{r\uparrow1}\frac{w\left(r\right)}{2\pi}\int_{0}^{2\pi}\left|f\left(re^{i\theta}\right)\right|d\theta<\infty\right\}$$

$$p_{w}\left(f\right)\overset{\textrm{def}}{=}\limsup_{r\uparrow1}\frac{w\left(r\right)}{2\pi}\int_{0}^{2\pi}\left|f\left(re^{i\theta}\right)\right|d\theta,\textrm{ }\forall f\in\mathcal{A}_{w}\left(\mathbb{D}\right)$$

Then, letting $f\in\mathcal{A}\left(\mathbb{D}\right)$ be arbitrary, define: $$W\left(f\right)\overset{\textrm{def}}{=}\left\{ w\in W:p_{w}\left(f\right)<\infty\right\}$$

Finally, fix $f\in\mathcal{A}\left(\mathbb{D}\right)$. I would be very happy if the condition: $$p_{w}\left(f\right)=0\textrm{ }\forall w\in W\left(f\right)$$

implied that: $$\limsup_{r\uparrow1}\left|f\left(r\right)\right|=0$$
or, even better: 

$$\sup_{\theta\in\mathbb{R}}\limsup_{r\uparrow1}\left|f\left(re^{i\theta}\right)\right|=0$$

(and hence, that $f$ is identically $0$ on $\mathbb{D}$, right?) but I don't know how to go about proving either of these. Any thoughts?